A system consists of two identical cubes, each of mass m, linked together by the compressed weightless spring of stiffness `ϰ` (figure). The cubes are also connected by a thread which is burned through at a certain moment. Find:
(a) at what values of `Deltal`, the initial compression of the spring, the lower cube will bounce up after the thread has been burned through:
(b) to what height h the centre of gravity of this system will rise if the initial compression of the spring `Deltal=7mg//ϰ`.

Explanation

(a) The initial compression in the spring `Deltal` must be such that after buring of the thread, the upper cube rises to a height that produces a tension in the spring that is atleast equal to the weight of the lower cube. Actually, the spring will first go from its compressed state to its natural length and then get elongated beyond this natural length. Let l bet the maximum elongation produced under these circumstances.
Then `kl=mg` (1)
Now, from energy conservation,
`1/2kDeltal^2=mg(Deltal+l)+1/2kl^2` (2)
(Because at maximum elongation of the spring, the speed of upper cube becomes zero)
From (1) and (2),
`Deltal^2-(2mgDeltal)/(k)-(3m^2g^2)/(k^2)=0` or, `Deltal=(3mg)/(k)`, `(-mg)/(k)`
Therefore, acceptable solution of `Deltal` equals `(3mg)/(k)`
(b) Let v the velocity of upper cube at the position (say, at C) when the lower block breaks off the floor, then from energy conservation.
`1/2mv^2=1/2k(Deltal^2-l^2)-mg(l+Deltal)`
(where `l=mg//k` and `Deltal=7(mg)/(k)`)
or, `v^2=32(mg^2)/(k)` (2)
At the position C, the velocity of C.M, `v_C=(mv+0)/(2m)=v/2`-Let, the C.M. of the system (spring+ two cubes) further rises up to `Deltay_(c2)`.
Now, from energy conservation,
`1/2(2m)v_C^2=(2m)gDeltar_(C2)`
or, `Deltay_(C2)=(v_C^2)/(2g)=(v^2)/(8g)=(4mg)/(k)`
But, uptil position C, the C.M. of the system has already elevated by,
`Deltay_(C1)=((Deltal+l)m+0)/(2m)=(4mg)/(k)`
Hence, the net displacement of the C.M. of the system, in upward direction
`Deltay_C=Deltay_(C1)+Deltay_(C2)=(8mg)/(k)`